Whakaoti i te x=sqrt{-3x+40}

$$x = \sqrt { - 3 x + 40 }$$

Whakaoti mō x

$$x=5$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$x^{2}=\left(\sqrt{-3x+40}\right)^{2}$$

Tātaihia te \sqrt{-3x+40} mā te pū o 2, kia riro ko -3x+40.

$$x^{2}=-3x+40$$

Me tāpiri te 3x ki ngā taha e rua.

$$x^{2}+3x=40$$

Tangohia te 40 mai i ngā taha e rua.

$$x^{2}+3x-40=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}+3x-40 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=-40$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -40.

$$-1,40 -2,20 -4,10 -5,8$$

Tātaihia te tapeke mō ia takirua.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=-5 b=8$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

$$\left(x-5\right)\left(x+8\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+8=0.

$$x=5 x=-8$$

Whakakapia te 5 mō te x i te whārite x=\sqrt{-3x+40}.

$$5=\sqrt{-3\times 5+40}$$

Whakarūnātia. Ko te uara x=5 kua ngata te whārite.

$$5=5$$

Whakakapia te -8 mō te x i te whārite x=\sqrt{-3x+40}.

$$-8=\sqrt{-3\left(-8\right)+40}$$

Whakarūnātia. Ko te uara x=-8 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$-8=8$$

Ko te whārite x=\sqrt{40-3x} he rongoā ahurei.

$$x=5$$